# Thread: Linear operators and Differential Equations

1. ## Linear operators and Differential Equations

Hello,

I have missed all my lectures on the above topic due to illness. Does anyone know of a good website with solved excercise sheets to help me learn material on this topic, in particular, I am looking for helpful notes for homework questions on this website http://www.maths.qmul.ac.uk/~cchu/MTH6122/lode092.pdf

Many Thanks

2. $\displaystyle <L(f),g>= \int_{0} ^{1} ( \int_{0} ^{1} f(y)K(x,y)dy )g(x)dx = \int_{0} ^{1} ( \int_{0} ^{1} g(x)f(y)K(x,y)dy )dx$ Using Fubini, this last one is equal to $\displaystyle \int_{0} ^{1} ( \int_{0} ^{1} g(x)f(y)K(x,y)dx )dy = \int_{0} ^{1} ( \int_{0} ^{1} g(x)K(x,y)dx )f(y)dy = <f,L(g)>$ So $\displaystyle L ^* =L$ ie. $\displaystyle L$ is self-adjoint.

Now suppose $\displaystyle f$ is an eigenfunction of $\displaystyle L$ with eigenvalue $\displaystyle \lambda$ with $\displaystyle K$ as defined in the sheet then:

$\displaystyle \lambda f(x) = \int_{0} ^{1} f(y)K(x,y)dy = \int_{0} ^{x} f(y)(1-x)ydy + \int_{x} ^{1} f(y)(1-y)xdy$$\displaystyle = (1-x) \int_{0} ^{x} yf(y)dy + x\int_{x} ^{1} (1-y)f(y)dy. In using this last expression for \displaystyle \lambda f(x) it's clear that \displaystyle f(0)=0=f(1) (as long as \displaystyle \lambda \neq 0, a difficulty that'll dissapear in the next step). Now, assuming \displaystyle f is twice differentiable (only continuity will not be enough) we have: \displaystyle \lambda f'(x)=-\int_{0} ^{x} yf(y)dy +(1-x)xf(x) + \int_{x}^{1} (1-y)f(y)dy - x(1-x)f(x)$$\displaystyle = -\int_{0}^{x} yf(y)dy +\int_{x}^{1} (1-y)f(y)dy$

$\displaystyle \lambda f''(x)=-xf(x) - (1-x)f(x)=-f(x)$ and so $\displaystyle \lambda f'' +f=0$. Now just notice that we never used $\displaystyle \lambda \neq 0$ so if $\displaystyle \lambda =0$ we just have $\displaystyle f=0$ which satisfies the boundary value problem trivially.

3. You are a life saver Jose! I will donate some money to this great site tonight.